$f\,^{\prime}(x)=\dfrac{16}{x^2}$ and $f(-2)=0$. $f(4) = $
Answer: Finding $f(x)$ We have $f'(x)=\dfrac{16}{x^2}$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int \dfrac{16}{x^2}\,dx \\\\ & = {-16x^{-1}} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(-2)=0$. Here's what we get when we plug in $-2$ : $\begin{aligned}f(-2)&={-16(-2)^{-1}} {+ C}\\\\ &={8} {+ C} \end{aligned}$ We are given that this must equal $0$ : $0 = {8} {+ C}$ Solving the equation gives us ${C=-8}$. Finding $f(4)$ Now, we have that $f(x)= {-16x^{-1}} {-8}$. Let's find $f(4)$ by plugging in $4$ : $\begin{aligned}f(4)&=-16(4)^{-1}-8\\\\ &=-12 \end{aligned}$ The answer $f(4) = -12$